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anonymous
 3 years ago
find the critical points: s(t)= (t1)^4 (t+5)^3
anonymous
 3 years ago
find the critical points: s(t)= (t1)^4 (t+5)^3

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first we find the derivative.. i know that.. then...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but wait, im confused to find the derivative for this one..

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0chain rule/product rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its 4(t1)^3 3(t+5)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what's next after it? i'm a little lost after it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.04(t1)^3 3(t+5)^2 = 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what about chain rule?

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.0doesnt apply here because derviative of t is just 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if ur function is of the type (t^2 + 2)^4 then s'(t) = 4 (t^2+2) * 2t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry s'(t) = 4 (t^2+2)^3 * 2t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if the derivative of t is 1 then how do i solve t?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry imm not sure what to do after ... 4(t1)^3 3(t+5)^2
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